Symplectic, Quaternionic, Fermionic - The Proof

John Baez

March 3, 2010

A) H has a conjugate-linear intertwining operator j with j² = 1, and H is equipped with an invariant orthogonal structure

or

B) H has a conjugate-linear intertwining operator j with j² = -1 and H is equipped with an invariant symplectic structure.

In the first case H comes from a real Hilbert space and j is called a "real structure". In the second case H becomes a quaternionic Hilbert space and j is called a "quaternionic structure".

Here's how we show this marvelous fact. The first step is to realize that the following are really equivalent things:

• a bilinear intertwining operator f: H × H → C
• a linear intertwining operator g: H ⊗ H → C
• a linear intertwining operator h: H → H^*
• a conjugate-linear intertwining operator j: H → H

hom(H,H^*)

This space is at most 1-dimensional by Schur's lemma, since we're assuming H and thus H^* are irreducible. And it's at least 1-dimensional since we're assuming there's an isomorphism between H and H^*. So it's 1-dimensional, all right.

But by the above stuff, this space is also isomorphic to

hom(H ⊗ H, C)

And we can write H ⊗ H as the direct sum of a space of symmetric tensors and a space of antisymmetric tensors:

H ⊗ H = S²H + Λ²H

so we have

hom(H ⊗ H, C) = hom(S²H,C) + hom(Λ²H,C)

Since the left-hand space is one-dimensional, one of the spaces on the right is one-dimensional, while the other is zero-dimensional. hom(S²H,C) is the space of invariant symmetric bilinear forms on H, while hom(Λ²H ,C) is the space of invariant skew-symmetric bilinear forms on H. So we either have one or other sort of form on H - not both. And tracing through the isomorphisms we see that these two cases yield two possibilities:

A) H has a conjugate-linear intertwining operator j with j² = 1, and H is equipped with an invariant orthogonal structure.

B) H has a conjugate-linear intertwining operator j with j² = -1 and H is equipped with an invariant symplectic structure.

If you didn't follow this, you can get more details in Adams' book "Lectures on Lie Groups".

Now you probably want to go back to where I talk about the implications of this marvelous fact.

One of the implications is that the intersection of U(2n) and Sp(2n,C) is isomorphic to Sp(n). Let me sketch the proof here. Say we have a complex Hilbert space with a symplectic structure. Then it must be even-dimensional, say 2n-dimensional. The above argument automatically gives it a quaternionic structure. The transformations that preserve the complex Hilbert space structure form the group U(2n). The transformations that preserve the symplectic structure form the group Sp(2n,C). The transformations that preserve both must automatically preserve the quaternionic structure as well, so they form the group Sp(n).

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